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I am confused on how objects are able to move, even though Newton's third law states that every action has an equal and opposite reaction.

I feel it would be easier to explain my misunderstanding in an example: If a horse is harnessed to a carriage, and applies a force on the carriage, the carriage applies a force back onto the horse. I understand that the "action" and "reaction" forces act on different objects, but wouldn't the "reaction" force applied back on the horse prevent the horse from moving? The horse applies a force onto the ground to start moving, but immediately after, this force transfers to the harness, and the harness pushes back onto the horse--no movement happens.

Qmechanic
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LottaQeustions
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    Does this answer your question? Given Newton's third law, why are things capable of moving? – Allure Feb 15 '23 at 01:19
  • The key mistake in your logic is the phrase: "this force transfers to the harness". No such thing happens. The static frictional force of the ground on the horse (forward) is independent of the force of the harness on the horse (backward). Imagine you are the horse: to get moving you press harder on the ground (backwards, you on ground), creating a larger Newton III pair on you (forwards, ground on you), until it exceeds the resistive force of the carriage pulling backwards on you. – Ben H Feb 15 '23 at 01:44
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    @Allure broken link, you must have linked to the post https://physics.stackexchange.com/questions/45653/given-newtons-third-law-why-are-things-capable-of-moving – nammerkage Feb 15 '23 at 08:34

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If the force applied to the horse by the carriage was equal to the force applied onto the ground by the horse, the horse would not move. But they are different if the carriage is accelerating.

To clear out the confusion, I recommend individually considering the free body diagrams for each object you're studying. You will immediately notice why things move the way they do.

The answer to the question why things accelerate is invariably that they have a net force acting upon them. (Philosophically, the relation is actually backwards, i.e. force is defined to be the rate of change in momentum.)

imcakmak
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As you noted, the key issue is that Newton's Third Law is a statement about forces on two different objects: $$ \vec{F}_\text{A on B} = - \vec{F}_\text{B on A} $$ Therefore you will never see these two forces within the same free-body diagram; they cannot "cancel out".

Draw the free-body diagram for the horse. There are: a tension force of the carriage backwards on the horse; a force of gravity downward on the horse; a normal force of the ground upward on the horse; and a static frictional force of the ground on the horse, pointing forward. If we state these forces with explicit reference to "object acting on object", they are: $$ \vec{F}_{T, \text{ carriage on horse}}\,, \, \, \vec{F}_{g, \text{ earth on horse}}\,, \, \, \vec{F}_{N, \text{ ground on horse}}\,, \, \, \vec{F}_{fs, \text{ ground on horse}} $$ Note that all of these forces belong on the free-body diagram of the horse, and are all "on horse". The acceleration of the horse will be determined, through Newton's Second Law, by the vector sum of these forces. In particular, for the horse to accelerate forward, the fourth of these forces must be larger in magnitude than the first. Those two forces are in no way related by Newton's Third Law.

The Newton's Third Law pairs of each of those forces are easy to write down (just reverse the "object on object", but maintain the type of force): $$ \vec{F}_{T, \text{ horse on carriage}}\,, \, \, \vec{F}_{g, \text{ horse on earth}}\,, \, \, \vec{F}_{N, \text{ horse on ground}}\,, \, \, \vec{F}_{fs, \text{ horse on ground}} $$ The first two forces belong in the free-body diagrams of the carriage and the (whole) earth, respectively, while the last two are on the free-body diagram of the ground. The acceleration of each of those objects will be due to the sum of forces in their own free-body diagrams. Again, Newton's Third Law plays no role in their dynamics.

The necessity of Newton's Third Law is to extend the dynamics of a single particle (which is prescribed by Newton's Second Law) to the dynamics of systems of particles. From this simple connection you can build up a theory the predicts the dynamics of rigid objects, clouds of particles interacting, fluids, and galaxies of stars.

Ben H
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but wouldn't the "reaction" force applied back on the horse prevent the horse from moving?

It depends on what other forces act on the horse.

If the static friction force between the horses feet and the ground equals the reaction force of the carriage on the horse, then the net force on the horse will be zero and the horse will not move per Newton's second law.

On the other hand, if the static friction force between the horse and the ground exceeds the maximum possible static friction force of $\mu_{s}Mg$, then the horse will accelerate in the direction of the reaction force of the carriage. An possible example is if the horse is standing on ice while the carriage is on dry ground.

Hope this helps.

Bob D
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  • I don't think this means anything: "if the static friction force between the horse and the ground exceeds the maximum possible static friction force of ". The horse accelerates forward if the forward-pointing static frictional force is larger than the backward-pointing force of the carriage on the horse. There is no need to invoke the maximum static frictional force here. – Ben H Feb 15 '23 at 01:03
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    Sorry, but I don't understand your point. The only thing opposing any horizontal movement of the horse in response to a horizontal force acting on the horse is static friction. Perhaps you can clarify your point. – Bob D Feb 15 '23 at 01:24
  • I'm only quarreling with the invocation of the maximum static frictional force. It is not relevant here. Moreover, the static frictional force cannot exceed the maximum static frictional force. I think maybe you just misspoke. – Ben H Feb 15 '23 at 01:25
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    But it is relevant. Static friction force equals the opposing force but only until the maximum possible static friction force is reached. If you don't like my answer that's OK and we can move on. – Bob D Feb 15 '23 at 01:29
  • Yes, that is true, but it is not relevant here. True: if a horse is going to walk, the static frictional force must be less than the maximum possible static frictional force (or else static friction will break and the horse will slip). But still the first sentence of your second paragraph doesn't make any sense. The issue is that the magnitude of the static frictional force (which, just in its existence, is something less than or equal to its maximum possible value) must be larger than the backwards force of the carriage on the horse, if the horse is to accelerate forward. – Ben H Feb 15 '23 at 01:33
  • And an "opposing force" is not always necessary for static friction. If you stand on ground and then launch yourself with a standing long jump, it is the static frictional force (ground on you) that launches you forward. But there is no "opposing force", and no need to invoke the maximum possible static frictional force to describe the motion. – Ben H Feb 15 '23 at 01:36
  • @BenH The static friction force the ground exerts forward on you acting on you is the reaction to your pushing backward on the ground. That's the opposing force I am referring to. – Bob D Feb 15 '23 at 02:02
  • I thought you were talking about, e.g., a box being pulled by a string. If there is static friction, then the harder you pull on the string, the larger the static frictional force on the box to keep it at rest... until your pulling force exceeds the max possible static frictional force, at which point it moves. I was just saying that this context of the static frictional force is not relevant here (the long jump example is closer to a relevant example). As to the static forces of "you on ground" and "ground on you" ... these are just a Newton's Third Law pair; they are always exactly equal. – Ben H Feb 15 '23 at 02:05
  • @BenH See my answer here: https://physics.stackexchange.com/questions/458258/can-you-jump-higher-if-you-run-if-so-why-high-jumping – Bob D Feb 15 '23 at 02:12
  • Yes, I have no argument with that, except that the max static frictional force is unnecessary (unless you are asking about whether the leaper is going to slip instead of leap). – Ben H Feb 15 '23 at 02:17